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The factorial $n!$ has a nice representation as definite integral: $$ n!=\Gamma(n+1)=\int_0^\infty t^{n} e^{-t}\, \mathrm{d}t. \! $$ Is it possible to write down such an integral for $n^n$ as well?

I tried to come up with an integral that reproduces a $n$ factor, $n$-times, but without success. I don't see a way to stop the partial integration process like in the $n!$ case. So this might not work here and I currently can't think of another way. If it helps to restrict $n$, feel free to do so.

The only thing a found online so far is the Lambert's $W$ function, which is involved when solving $x^x=z$, but I'm not sure if this helps.

EDIT: Answers with integrals of the form $\displaystyle n^n=\int_0^\infty \cdots dt$ are preferred.

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  • $\begingroup$ Asymptotics like $\displaystyle n^n\approx \int_0^\infty (2\pi n)^{-1}(et)^{n} e^{-t}\, \mathrm{d}t $ don't count. $\endgroup$
    – draks ...
    Jul 26, 2012 at 8:47

4 Answers 4

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No need for the Lambert W function , respecting your limits and using $\Gamma$ we get that

$$ n^{n}=\frac{1}{\Gamma (n+1)}\int_{0}^{\infty }e^{-\frac{t^{ \frac{1}{n}}}{n}} dt $$

Update:

Lookup the Exponential Integral and its relationship with the Incomplete gamma function

$$E_{n}(x)=x^{n-1}\Gamma(1-n,x)\tag{1}$$

for $n=1-n$ and $x=\frac{1}{n}$ we get that :

$$E_{1-n}(\frac{1}{n})=\int_{1}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt = n^{n}\Gamma (n,\frac{1}{n})\tag{2}$$

Changing the integration limits to $[0,1]$ :

$$\int_{0}^{1}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt = n^{n}(\Gamma(n)-\Gamma (n,\frac{1}{n}))\tag{3}$$

Combining 2+3 we get the solution:

$$n^{n}=\frac{1}{\Gamma(n)}\int_{0}^{\infty}\frac{e^{-\frac{t}{n}}}{t^{1-n}}dt$$

The top result is due to the relationship :

$$n E_{1-n}(\frac{1}{n})= \int_{1}^{\infty} e^{-\frac{t^{\frac{1}{n}}}{n}} dt\tag{4}$$

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  • $\begingroup$ Thanks again, how did you get that? $\endgroup$
    – draks ...
    Jul 27, 2012 at 17:10
  • $\begingroup$ @draks update how $\endgroup$
    – GerryMrt
    Jul 30, 2012 at 12:15
  • $\begingroup$ Thanks again, great answer, vote up people$\color{green}{.}\color{goldenrod}{.}\color{red}{.}$ $\endgroup$
    – draks ...
    Jul 30, 2012 at 12:19
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This may not be the line of thinking you are after, but it does give the integral you desire.

$$n^n=\int_0^n nx^{n-1}\ dx$$

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  • $\begingroup$ Thanks, no it's not. I thought about an infinite upper limit. But despite that it looks very interesting. +1 $\endgroup$
    – draks ...
    Jul 26, 2012 at 11:02
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$$\frac{\Gamma (\alpha)}{s^{\alpha}}=\int_{0}^{\infty }t^{\alpha-1}e^{-st}dt\tag{1}$$

$$\frac{\Gamma (s)}{s^{s}}=\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{2}$$

$$s^{-s}=\frac{1}{\Gamma (s)}\int_{0}^{\infty }t^{s-1}e^{-st}dt\tag{3}$$

$$s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}\tag{4}$$

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  • $\begingroup$ Hmm, not bad +1. invert it and we are even closer $\displaystyle s^{s}=\frac{\Gamma (s)}{\int_{0}^{\infty }t^{s-1}e^{-st}dt}$, at least on the lhs... $\endgroup$
    – draks ...
    Jul 26, 2012 at 11:05
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The correct question (which someone has just asked me) may be as follows:

Find $f(t)$ (positive and independent of $n$) such that

$$n^n = \int_0^\infty t^n f(t) \,dt,\quad \forall\,n$$

That is, find $f(t)$ such that $\{n^n\}$ is the sequence of moments of f(t)dt. This is a particular instance of the Stieltjes moment problem.

There is criterion (due to Carleman) which guarantees that such a solution is unique, but I don't know an explicit form for $f(t)$.

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    $\begingroup$ A rather strange edit of this post has been approved. However, it is possible that user65120 and the anonymous user, who suggested the edit, are the same person. $\endgroup$ Mar 5, 2013 at 12:30
  • $\begingroup$ thanks for the background infos... $\endgroup$
    – draks ...
    Mar 5, 2013 at 14:28

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